Question: Let $R$ be the region to the right of the line $x=1,$ and enclosed by it, the line $y=4$, and the curve $y=(x-1)^2$. $y$ $x$ ${y=(x-1)^2}$ ${y=4}$ $x=3}$ $(1,0)$ $(3,4)$ $ R$ A solid is generated by rotating $R$ about the line $x=3$. What is the volume of the solid? Give an exact answer in terms of $\pi$.
Explanation: Let's imagine the solid is made out of many thin slices. Each slice is a cylinder with a hole in the middle, much like a washer. $y$ $x$ ${y=(x-1)^2}$ ${y=4}$ $x=3}$ $(1,0)$ $(3,4)$ Let the thickness of each slice be $dy$, let the radius of the washer, as a function of $y$, be $r_1(y)$, and let the radius of the hole, as a function of $y$, be $r_2(y)$. Then, the volume of each slice is $\pi[(r_1(y))^2-(r_2(y))^2]\,dy$, and we can sum the volumes of infinitely many such slices with an infinitely small thickness using a definite integral: $\int_a^b \pi [(r_1(y))^2-(r_2(y))^2]\,dy$ This is called the washer method. What we now need is to figure out the expressions of $r_1(y)$ and $r_2(y)$, and the interval of integration. $r_1(y)$ is equal to the distance between the line $x=1$ and the line $x=3$. So, ${r_1(y)=2}$. $r_2(y)$ is equal to the distance between the curve $y=(x-1)^2$ and the line $x=3$. To find it, we need to solve the equation for $x$ : $x=\sqrt y+1$ So, ${r_2(y)=2-\sqrt y}$. Now we can find an expression for the area of the washer's base: $\begin{aligned} &\phantom{=} \pi [({r_1(y)})^2-({r_2(y)})^2] \\\\ &= \pi [({2})^2-({2-\sqrt y})^2] \\\\ &=\pi\left[ 4-(4-4\sqrt y+y) \right] \\\\ &=\pi(-y+4\sqrt y) \end{aligned}$ The bottom endpoint of $R$ is at $y=0$ and the top endpoint is at $y=4$. So the interval of integration is $[0,4]$. Now we can express the definite integral in its entirety! $\begin{aligned} &\phantom{=}\int_0^4 \left[\pi(-y+4\sqrt y) \right] dy \\\\ &=\pi\int_0^4 (-y+4\sqrt y)dy \end{aligned}$ Let's evaluate the integral. $\pi\int_0^4 (-y+4\sqrt y)dy=\dfrac{40\pi}{3}$ In conclusion, the volume of the solid is $\dfrac{40\pi}{3}$.